The frequency `f` of vibrations of a mass `m` suspended from a spring of spring constant `k` is given by `f = Cm^(x) k^(y)` , where `C` is a dimensionnless constant. The values of `x and y` are, respectively,

Options

A. `x=1/2,y=1/2`
B. `x=(-1)/2,y=(-1)/2`
C. `x=(1)/2,y=(-1)/2`
D. `x=(-1)/2,y=(1)/2`

Answer

To solve the problem, we need to determine the values of \( x \) and \( y \) in the equation for frequency \( f \) given by:

\[ f = C m^{x} k^{y} \]

where \( C \) is a dimensionless constant, \( m \) is the mass, and \( k \) is the spring constant.

### Step 1: Identify the dimensions of each variable

1. **Frequency \( f \)** has dimensions of time inverse:
\[ [f] = T^{-1} \]

2. **Mass \( m \)** has dimensions:
\[ [m] = M \]

3. **Spring constant \( k \)** has dimensions derived from Hooke's law, where the force \( F \) is given by:
\[ F = kx \]
Here, \( F \) has dimensions of force, which is:
\[ [F] = M L T^{-2} \]
and \( x \) (displacement) has dimensions of length:
\[ [x] = L \]
Therefore, rearranging gives:
\[ [k] = \frac{[F]}{[x]} = \frac{M L T^{-2}}{L} = M T^{-2} \]

### Step 2: Write the dimensional equation

Now we can substitute the dimensions into the equation:
\[ [T^{-1}] = [m^{x}][k^{y}] \]
Substituting the dimensions we found:
\[ [T^{-1}] = [M^{x}][(M T^{-2})^{y}] \]

### Step 3: Expand the right-hand side

Expanding the right-hand side:
\[ [T^{-1}] = M^{x + y} T^{-2y} \]

### Step 4: Equate the dimensions

Now we equate the dimensions on both sides:
1. For mass \( M \):
\[ x + y = 0 \]
2. For time \( T \):
\[ -2y = -1 \implies y = \frac{1}{2} \]

### Step 5: Solve for \( x \)

Using the value of \( y \) in the first equation:
\[ x + \frac{1}{2} = 0 \implies x = -\frac{1}{2} \]

### Conclusion

Thus, the values of \( x \) and \( y \) are:
\[ x = -\frac{1}{2}, \quad y = \frac{1}{2} \]

### Final Answer

The values of \( x \) and \( y \) are respectively:
\[ \boxed{-\frac{1}{2}} \quad \text{and} \quad \boxed{\frac{1}{2}} \]

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